What are the inflection points for #y = 6x^3 - 3x^4#?

Question

William Abdalla · Accepted Answer

#(0,0), (1,3)#

Determine the initial derivative:

#y'=18 x^2-12 x^3#

Calculate the second derivative:

#y''=36x-36 x^2#

Let's factor #y''# a bit to simplify it:

#y''=36x(1-x)#

Set #y''# equal to zero and solve for #x#:

#36x(1-x)=0#

#36x=0 #

So #x=0# is our first potential inflection point's #x#-value.

#1-x=0#

So #x=1# is our second potential inflection point's #x#-value.

We must test values of #y''# in the following intervals (but not at the endpoints, hence the parentheses):

#(-∞, 0)# #(0, 1)# #(1, ∞)#

#(-∞,0)#

#y''(-1)=-36(2) <0 #

In the interval #(-∞, 0#), #y'' < 0# and so the graph of y is concave down.

#(0,1)#

#y''(1/2)=18(1/2) >0 #

In the interval #(0,1)#, #y''>0# and so the graph of y is concave up. We've switched concavity. This means we have an inflection point at #x=0#.

Let's plug #x=0# back into our original function to get the inflection point's coordinates:

#y(0)=6(0)^3-3(0)^4=0#

#(0,0)# is an inflection point.

#(1,∞)#

#y''(2)=72(-1)<0 #

In the interval #(1,∞)#, #y''<0# and so the graph of y is concave down. Again, we've switched concavity. We also have an inflection point at #x=1#.

Plug #x=1# into the original function to get the second inflection point's coordinates:

#y(1)=6-3=3#

#(1,3)# is an inflection point.

Evelyn Agard · Answer

undefined To find the inflection points of $ y = 6x^3 - 3x^4 $, we need to find where the second derivative changes sign.

First derivative:
$$ y' = \frac{d}{dx}(6x^3 - 3x^4) $$

$$ y' = 18x^2 - 12x^3 $$

Second derivative:
$$ y'' = \frac{d}{dx}(18x^2 - 12x^3) $$

$$ y'' = 36x - 36x^2 $$

To find the inflection points, set the second derivative equal to zero and solve for $ x $:
$$ 36x - 36x^2 = 0 $$

$$ 36x(1 - x) = 0 $$

$$ x = 0, \, x = 1 $$

So, the inflection points are $ x = 0 $ and $ x = 1 $.